GATE_PROJ
Top 16 Positive Activations
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8 K (25°C) yields ~168.5 m/s. Therefore, since the
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and rounded to three sig figs is 169 m/s. Thus, present answer boxed as
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compute", so it's expected to compute.\n\nAlternatively, perhaps check if there is an established average speed for
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precision needed.\n\nAlternatively, maybe the original question assumes a specific temperature but forgot to mention. If the user
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and calculator computations, perhaps the answer expects 170 m/s as an approximate answer. But some
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98 K (25°C) yields ~168.5 m/s. Therefore, since
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. Since the user hasn't specified temperature, perhaps there's a standard temperature to assume. Let me check
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different.\n\nBut given that the user hasn't specified the temperature, this seems ambiguous. However, in many
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them as a, b, c such that one is the product of the other two. But maybe no
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is rounded to three significant figures, giving \(169 \, \text{m/s}\).\n\n
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around 298 K. Maybe. Alternatively, as room temperature sometimes is taken as 30
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's expected to compute.\n\nAlternatively, perhaps check if there is an established average speed for radon gas at
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, given that no temperature is given, perhaps they just want the formula, but I don't think so
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, if forced to choose, answer approximately 170 m/s. But given compute precisely, it
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=1, perhaps consider that one of the variables is the product of conjugates of the others. For
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m/s. But if the question hasn't specified the temperature, this can vary.\n\nAlternatively, given that
Top 16 Negative Activations
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3000 → 0.57 * 6000 = ?\n\nWait, maybe
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0.57 * 6000 = ?\n\nWait, maybe better to process 60
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Let's see.\n\n168^2 = (170 - 2)^2 =
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2: 1*1 = 1, 1*0.12 = 0.
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0×30 =7200, 7440-7200=
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2.\n\nWait, let me check that subtraction:\n\n9,628.24\n\n-9
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squared. Let's break it down:\n\nFirst, 1 * 1.2544 =
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x1.12\n--------\n224 (1.12 * 2)\n11
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,030,000, since 5025*1000=
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24=14,328 and 597*0.8004
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me check that subtraction:\n\n9,628.24\n\n-9,441.
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000 = 6000\n\n0.5 * 6000 =
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11616 = ?\n\nCompute that:\n\n9628.23864 -
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,441.12\n\n------------\n\nSubtract dollars: 9628 -
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0. Let's see: 240×30 =7200, 7
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by 240. Let's see: 240×30 =720
UP_PROJ
Top 16 Positive Activations
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but their exponents are 1, which is odd. Actually, wait nvm, the exponents
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). The altitude is h, the horizontal distance from G to EM. So h is the length we need
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7 terms) is-1 mod1000. So multiplying 891 by (-1
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win. Wait, 2 is a losing position, so if first player moves from 3 by taking
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:4,7:2, which are all even? Wait, 18,8,4
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, so 0.88% is already two significant figures. Alternatively, if written as 0
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= sqrt(2) * ||x - y|| / sqrt{1 - a^, but this
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care. So GEM is an isosceles triangle with base EM, so EM is from (
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, next state (t=1): O T O.\n\nNow, check each lamp:\n\n- L1
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128\), since each pair is counted twice. But actually, since we are considering ordered pairs
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will turn off. If it's off, it stays off. Therefore, regardless of the starting state (
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sum 0 or 2, it's 0. Therefore, XOR gives the same result as modulo
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exponent of 7 is 2, which is even. Wait a second. Then primes 11
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- y) : x ∈ S } are orthonormal. Hence, they can be considered as an
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the vectors from the centroid to the vertices are orthonormal. Wait, not exactly, but in the
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each unordered pair {a, b} is counted twice in the 256 ordered pairs.\n\nBut
Top 16 Negative Activations
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infinitely many positions exist. As a contest problem, the solution may be found in realizing the periodicity failed
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time.\n\nSo perhaps the answer is not straightforward, hence why reference to the P/poly theorem. The
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Either way, this isn't helping. Perhaps I need a different approach.\n\nAlternative idea: Use Hölder
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the possible patterns and their evolutions. Perhaps a solution exists where nが is a power of two?
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, power configurations may make difference.\n\nAlternatively, perhaps answer is that n must be a power of2
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the person who set the problem must have an particular method in mind. Let's go back.\n\nAnother strategy
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in circles here. Given the time constraints, maybe for problem 3, since it's known in complexity
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who set the problem must have an particular method in mind. Let's go back.\n\nAnother strategy. Use
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exist. As a contest problem, the solution may be found in realizing the periodicity failed, but that
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only countable ones.\n\nBut the question is probably supposed to allow the System S to be mapped to or
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P/poly.\n\nTherefore, for the first problem, combining multiple sparse sets into one via tagging. For
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positions exist. As a contest problem, the solution may be found in realizing the periodicity failed, but
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person who set the problem must have an particular method in mind. Let's go back.\n\nAnother strategy.
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the subject.\n\nAlternatively, but, finally, the answer is asserting that such a y exists because he’s
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. As a contest problem, the solution may be found in realizing the periodicity failed, but that a
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problem asks to show it, not just cite, perhaps construction:\n\nGiven L ∈ P_angel, then
DOWN_PROJ
Top 16 Positive Activations
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1, 1/a = bc, 1/b = ac, 1/c = ab, as
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Hmm. So q = a^2 c + b^2 a + c^2 b. Since
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sqrt(0.000064) is 0.008, and sqrt
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. Therefore, q = a^2 c + b^2 a + c^2 b. Hmm
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=30,150. Then, 30,150 /502
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can write 1/a = bc, 1/b = ac, 1/c = ab. So
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= ab. So 1/a + 1/b + 1/c = bc + ac + ab
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. But then abc = (1/b)(1/c)(1/a) = 1/(a b
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is (a + 1/a) + (b + 1/b) + (c +
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0, since 5025*1000=5,025,0
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0.\n\nFirst, multiply 6000 by 1.5735193
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is not 1. So all of a, b, c are complex but non-real, with modulus
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Then indeed, if a = 1/b, b = 1/c, c = 1/a
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14,328 and 597*0.8004≈ 5
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0.57351936\n\nWait, or perhaps it's easier to multiply
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1=0, but they are 1 and the complex roots. But those have modulus 1.
Top 16 Negative Activations
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1200. Yep, that's right. So, C_eq is 120
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between 3 and 2? Let's see:\n\n0 to2: +2\n\n2 to5
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,0)):\n\nThe slope (m) is (0 -5)/(10 - (10
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{w(x)} f(x) dx \leq \left( \int_0^\infty
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8≈168, which is about right. So, the number looks consistent. Therefore, if
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respectively. Then, error terms calculated for each. \n\nAlternatively, perhaps I could use logarithmic differentiation.
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..., up to k=999 (10^999 -1). Therefore, the
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,2) = same 60*20= 1200. Times 2
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Let's see:\n\n0 to2: +2\n\n2 to5:+3\n\n5 to7:+
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1}=2→Grundy[19}=2\n\nn=20: subtract1→1
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0 - (10 - h)) = (-5)/h.\n\nThe equation is y -0 =
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5*4)=2*60*20=2400\n\nk=4:\n\n
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sequences: R GGGG, GGGG R\n\nk = 2: RR GGG,
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λ(x) = λ^{a} f(λ x). Then:\n\nCompute I_λ = �
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moves to W. So n=20:L.\n\nn=21: n -1=2
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, etc., would give the same result as before.\n\nLet me actually compute it this way:\n\nFirst term